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Module 13

Getting a feel for the path integral

Build the picture that every path contributes, and watch the classical path emerge from the interference.

What you will see

  • Hundreds of ragged paths connecting the same two points, each carrying a phase
  • Phase encoded as colour: far from the classical path the colour spins fast and contributions cancel
  • Shrink ħ and only a thin bundle around the classical path survives

Assumed background

  • Complex phases and interference (module 01)
  • Action and the principle of least action in classical mechanics (helpful)

The Schrödinger equation tells you how the wavefunction evolves. Feynman gave a second account of the same physics, and it sounds far more outrageous:

To go from A to B, the particle takes every possible path. Each path contributes an amplitude eiS/ℏ\ee^{\ii S/\hbar}, and the total amplitude is their sum.

Every path — including the one that loops around the Earth first, the one that doubles back, the one whose velocity jumps.

The path integral: every path contributes, the classical one is merely the survivor

Each path carries a phase e^{iS/ħ}, and the colour *is* that phase. Chaining all the amplitudes head to tail gives the strand at the top — wherever it curls into a spiral, paths are cancelling each other.

Loading 3D scene…

ħ
1.00

This is the only switch in the scene. The smaller ħ, the larger the phase difference for the same ΔS, the more complete the cancellation, and the narrower the bundle of surviving paths.

Path sample
120
1.10

How far the random paths wander from the straight line

Display
K(b;a)=∫D[x(t)]  eiS[x]/ℏ,S[x]=∫L dtK(b;a) = \int \mathcal{D}[x(t)]\;\ee^{\ii S[x]/\hbar},\qquad S[x]=\int L\,\dd t
Classical action S_cl = 23.120ħ = 1.00Phase sensitivity S_cl/ħ = 23.1 rad38% of paths have |ΔS| < πħTotal amplitude |ΣA|/N = 0.210
  • classical path (S stationary)
  • Phasor chain
  • path colour = phase (colour wheel)

What to look for

  • Start with ħ = 3 (quantum): the colours change gently, meaning neighbouring paths differ little in phase and all of them contribute effectively. The phasor chain is nearly straight — everything adds in phase.
  • Drag ħ down to 0.1 (the classical limit): the colours start spinning fast and the chain immediately curls into a tight spiral going nowhere. Only the paths hugging the classical one still add in phase.
  • Watch the fraction of paths with |ΔS| < πħ in the readout: the smaller ħ, the lower it goes. That surviving bundle *is* the width of the classical trajectory.
  • Increase the path roughness: wilder paths appear, their phases are more scrambled and their net contribution is almost nothing — but they do take part.
  • Switch to the phasor view and face the spiral: the straight line from the start of the chain to its end is the total amplitude. The straighter the chain, the larger it is.

✕ "The particle really does travel every path at once"

The path integral is a way of computing an amplitude, not a literal account of what the particle did. The safe statement is: when the path is not specified, all path amplitudes must be summed.

✕ "Paths far from the classical one contribute little"

Every path has exactly the same amplitude *magnitude*. Only the phases differ. Distant paths do not contribute little — they cancel each other. The two statements build completely different intuitions.

✕ "Typical paths are smooth"

The dominant paths are continuous everywhere and differentiable nowhere (⟨Δx²⟩ ∝ Δt, exactly as in Brownian motion). "Velocity" is not a well-defined quantity in this framework.

Think it through

  1. In classical mechanics the principle of least action is a postulate. What is it in the path integral? (Hint: watch what the phasor chain does as ħ → 0.)
  2. How would you describe the double slit in this language? (Hint: the barrier kills almost every path and leaves two families.)
  3. If a path has S larger than the classical value by 100ħ, "being cancelled" means some other path is exactly out of phase with it. Does that pairing still exist when ħ is large?

The propagator is a sum

K(xb,tb;xa,ta)=∫D[x(t)] eiS[x]/ℏ,S[x]=∫tatbL(x,x˙) dt.K(x_b,t_b;x_a,t_a) = \int \mathcal{D}[x(t)]\ \ee^{\ii S[x]/\hbar}, \qquad S[x] = \int_{t_a}^{t_b} L(x,\dot x)\,\dd t .

Every path has exactly the same amplitude magnitude; only the phase differs. That is deeply counter-intuitive: an absurd path contributes as much amplitude as the classical one.

So what makes the classical path special?

Stationary phase: cancelling and not cancelling

Near the classical path the action SS is stationary, meaning its first variation vanishes: neighbouring paths have nearly the same phase, so their amplitudes add in phase and reinforce.

Far from it, SS changes rapidly, neighbouring phases are scrambled, and the contributions cancel.

The scene encodes phase as colour. You will see a band around the classical path where the colour changes gently, and beyond it the colour spinning into noise. Almost all of the sum comes from that band.

Why the formulation is worth learning

The path integral is fully equivalent to the Schrödinger equation (you can derive one from the other), but it wins outright in several places:

  • Field theory: quantising gauge fields and deriving Feynman rules is far more natural this way;
  • Statistical mechanics: substitute t→−iτt\to-\ii\tau and the path integral becomes a partition function, making quantum mechanics and statistical physics the same mathematics;
  • Topological effects: Aharonov–Bohm phases, instantons and θ-vacua are cleanest when you sum by topological class;
  • Numerics: quantum Monte Carlo samples directly in path space.
From the path integral back to the Schrödinger equationoptional

For an infinitesimal time ϵ\epsilon, the single-step propagator is

K(x′,t+ϵ;x,t)≈m2πiℏϵexp⁡{iℏ[m(x′−x)22ϵ−ϵV ⁣(x+x′2)]}.K(x',t+\epsilon;x,t) \approx \sqrt{\frac{m}{2\pi\ii\hbar\epsilon}} \exp\left\{\frac{\ii}{\hbar}\left[\frac{m(x'-x)^2}{2\epsilon} - \epsilon V\!\left(\tfrac{x+x'}{2}\right)\right]\right\}.

Substituting into ψ(x′,t+ϵ)=∫K ψ(x,t)dx\psi(x',t+\epsilon)=\int K\,\psi(x,t)\dd x, writing η=x′−x\eta = x'-x and expanding to second order, the Gaussian integrals give

⟨η2⟩=iℏϵm,\langle\eta^2\rangle = \frac{\ii\hbar\epsilon}{m},

so that

ψ(x,t+ϵ)=ψ+iℏϵ2m∂x2ψ−iϵℏVψ+O(ϵ2),\psi(x,t+\epsilon) = \psi + \frac{\ii\hbar\epsilon}{2m}\partial_x^2\psi - \frac{\ii\epsilon}{\hbar}V\psi + O(\epsilon^2),

which rearranges into

iℏ ∂tψ=−ℏ22m∂x2ψ+Vψ.\ii\hbar\,\partial_t\psi = -\frac{\hbar^2}{2m}\partial_x^2\psi + V\psi .

Note that ⟨η2⟩∝ϵ\langle\eta^2\rangle \propto \epsilon rather than ϵ2\epsilon^2: the typical path is continuous everywhere and differentiable nowhere, just like Brownian motion. “Velocity” is simply not a well-defined quantity in this framework.

Think it through

  1. How would you describe the double slit in this language? (Hint: the barrier kills almost every path and leaves two families.)
  2. If a path has an action 100ℏ100\hbar larger than the classical one, is its contribution “small” or “cancelled”? What is the difference between those two statements?
  3. Typical paths are nowhere differentiable. What, then, is “the velocity of the particle” in the path integral, and how does that connect to the uncertainty principle?

Go deeper · matching textbook sections

The 3D scenes build the picture; the full derivations and exercises live in the textbook.

Having finished this module